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The (normalized) Ricci flow on compact surfaces is given by $$\frac{\partial}{\partial t}g_{ij}=(r-R)g_{ij}\text{ ,}$$ and in the beginning of Hamilton's paper on the topic he points out that since the rate of change of the metric is, pointwise, a multiple of the metric, the metric is flowed within its conformal class. That is, if the initial metric is $g_0$, the metric at later times would be something like $$g(t)=\varphi g_0$$ where $\varphi$ is some scalar function. Intuitively, this explanation makes sense, but is there a rigorous way of showing this?

For an ODE of the form $f'(x)=h(x)f(x)$ we would expect the answer to be $f_0$ scaled by some exponential. Is the reasoning similar in this case? Since the scalar curvature $R$ depends implicitly on the changing metric $g(t)$, I expect this would complicate things.

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The dependence of the curvature on $g$ is somewhat of a red herring here - it certainly makes solving the equation difficult, but assuming we have a smooth solution of Ricci flow defined on $M \times [0,T)$, the quantity $f(x,t) = r - R_{g(t)}(x)$ is a well-defined smooth function on $M \times [0,T)$.

Thus the Ricci flow equation becomes $$\frac{\partial}{\partial t}\ln g_{ij}(x,t) = f(x,t),$$ so integrating we get $$\ln g_{ij}(x,t) = \ln g_{ij}(x,0)+\int_0^t f(x,s) ds;$$

i.e. $$g_{ij}(x,t) = g_{ij}(x,0)\exp\left(\int_0^t f(x,s) ds\right).$$

If you already believe that solutions to the flow are unique, another way to view this is as an ansatz - if you assume $g(t) = \varphi\cdot g(0)$ and plug-and-chug the formula for the behaviour of the curvature under conformal transformations, you should find that such a $g$ satisfies the Ricci flow when $\varphi$ satisfies a scalar heat-type equation of the form $$\partial_t \varphi = \Delta_{g_0} \ln \varphi + r \varphi - R_{g_0}.$$ Since this is a strongly parabolic equation, it has a solution with initial condition $\varphi(x,0) = 1$, which gives rise to a solution of Ricci flow of the desired form.

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